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Need help with getimagesize

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dreammaker888

dreammaker888

Programmer
Jul 31, 2007
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I have the following code but $ImageSize returns empty.
What am I doing wrong?

Code: 
$result = mysql_query($qdf)
	or die("<BR>Invalid query: $qdf <BR>" . mysql_error() . "".$result);

	$row = mysql_fetch_array($result);

	$pic = $row['PicURL'];
	$ImageSize = getimagesize($pic);


	echo $ImageSize[0];
	echo $ImageSize[1];	
	echo $ImageSize[2];
	echo $ImageSize[3];

Do I have to declared $ImageSize as array somewhere?

What exactly do each of the array should return 0, 1, 2, etc.? I am looking for the height and the width of the picture.

Thanks.
 
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The array should contain something similar to the following if it can get the information from the file.

Code: 
Array
(
    [0] => 60  [green]\\Width[/green]
    [1] => 60  [green]\\Height[/green]
    [2] => 1   [green]\\Image Type[/green]
    [3] => width="60" height="60" [green]\\String for HTML Image tag[/green]
    [bits] => 8 
    [channels] => 3
    [mime] => image/gif 
)

If its not retuning anything, perhaps it can't access the image.

What is in your $pic variable?

Is your $ImageSize variable returning FALSE?
You can check it like this:
Code: 
if($ImageSize===FALSE)
{
echo "Its False";
}



----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.
 
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You don't need to declare it. By just using the the function it should get declared and contain the values I posted.

So this is fine: $ImageSize = getimagesize($pic);

If you want to declare it:
$ImageSize=array(); should be enough.

Running getimagesize on the image URL you provided yields the following information for it:

Code: 
Array
(
    [0] => 576
    [1] => 527
    [2] => 2
    [3] => width="576" height="527"
    [bits] => 8
    [channels] => 3
    [mime] => image/jpeg
)

So there must be something else wrong there. Now You may be getting an error you are unaware of.

Can you take a look at PHP's error log see if it has anythng relevant?

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.
 
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Can you post the complete code that got you those result?

Thanks.

 
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Sure, its nothing fancy:

Code: 
$pic="[URL unfurl="true"]http://home.att.net/~DMK2/TriHead.jpg";[/URL]
$ImageSize=getimagesize($pic);
echo "<pre>";
print_r($ImageSize);
echo "</pre>";



----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.
 
Upvote 0 Downvote
you need to make sure that your system is set up to allow you to access foreign url's. it's in the php.ini file.
 
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I put this in a php file and when I ran it it returned nothing.

What could be wrong?

Code: 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "[URL unfurl="true"]http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">[/URL]
<html xmlns="[URL unfurl="true"]http://www.w3.org/1999/xhtml">[/URL]
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php

echo "xx<br>";
$pic="[URL unfurl="true"]http://home.att.net/~DMK2/TriHead.jpg";[/URL]
$ImageSize=getimagesize($pic);
echo "<pre>";
print_r($ImageSize);
echo "</pre>";

echo "yy";
?>
</body>
</html>
 
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I may know why? It could be the encoding of the "~" in front of DMK.

The space is %20, what is the encoding for ~?

 
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Actually, my hosting is having problem with the getimagesize function when the jpg file is residing in another server (in this case at att.net).

If I copy the file into the host server, the function worked and returned the proper array.

The problem I have is, it is impractical for me to copy the thousands of pictures from att.net onto my host.

Does anyone know how to solve this delimma?

 
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Sorry, missed your post in the haste of testings.

This is new to me. Where is php.ini residing?
 
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check the settings through a call to phpinfo()
 
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As Jpadie suggested use a call to phpinfo(), and look for the entry for allow_url_fopen.

It must be set to on.

If its not set to on, you won't have access to remote urls.



----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.
 
Upvote 0 Downvote
but you can try a local php.ini file. just create a text file with the relevant directive in it. save it as php.ini and upload it to the script directory in question. sometimes an ISP will allow overrides like these.
 
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OK. I've got it.

Modified the ini file and everything works like charm.

Thank you, thank you, THANK YOU.

 
Upvote 0 Downvote
i suspect you would have got their earlier if your error display had been turned on. for debugging i would recommend that you include these lines at the start of your script
Code: 
ini_set('error_display', 'ON');
error_reporting(E_ALL);

or change the relevant settings in php.ini
 
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